Undergrad Scalar product and vector product

Click For Summary
SUMMARY

The discussion focuses on the mathematical concepts of the cross product and dot product of vectors A and B. The cross product, defined as A X B, results in a vector that is perpendicular to the plane containing A and B, fulfilling the requirement of being anti-commutative. The proof for the dot product formula A.B = |A||B|cos(θ) is derived using the cosine rules for vector addition and subtraction, establishing the relationship between the magnitudes of the vectors and the cosine of the angle between them.

PREREQUISITES
  • Understanding of vector operations, specifically cross product and dot product.
  • Familiarity with trigonometric identities and cosine rules.
  • Basic knowledge of Euclidean space and inner product spaces.
  • Concept of linear combinations and anti-commutativity in vector mathematics.
NEXT STEPS
  • Study the properties of vector products, focusing on the geometric interpretation of the cross product.
  • Explore the derivation of the dot product formula in various contexts, including inner product spaces.
  • Learn about the applications of vector products in physics, particularly in rotational dynamics.
  • Investigate the implications of anti-commutativity in vector algebra and its applications in higher dimensions.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are looking to deepen their understanding of vector operations and their applications in various fields.

prashant singh
Messages
56
Reaction score
2
why do we take cross product of A X B as a line normal to the plane which contains A and B. I also need a proof of A.B = |A||B|cos(theta), I have seen many proves but they have used inter product ,A.A = |A|^2, which is a result of dot product with angle = 0, we can't use this too prove the dot product formula.
First one is more important please help.
 
Mathematics news on Phys.org
prashant singh said:
why do we take cross product of A X B as a line normal to the plane which contains A and B.
You asked why, well it's because a vector product is defined that way. It's defined such that the result of the product is perpendicular to any linear combination of A and B.
prashant singh said:
I also need a proof of A.B = |A||B|cos(theta)
Consider the cosine rules for the addition and the difference between two vectors.
$$
|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos\theta \\
|\vec{A}-\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos\theta \\
$$
 
Regarding the first question: We need some fundamental operation that reflects rotations in R3 from direction A to direction B. A nice operation would be linear in both A and B and would be anti-commutative ( AxB = -BxA ). The definition of the cross product fits the bill.
 
In euclidean 2D space, if \mathbf{a} = (a \cos \phi, a \sin \phi) and \mathbf{b} = (b \cos \alpha, b \sin \alpha) then by basic trigonometry \mathbf{a} \cdot \mathbf{b} = (a \cos \phi, a \sin \phi) \cdot (b \cos \alpha, b \sin \alpha) = ab \cos \phi \cos\alpha + ab \sin \phi \sin \alpha = ab \cos(\phi - \alpha), or <br /> \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\| \cos\theta where \theta = \phi - \alpha is the angle between \mathbf{a} and \mathbf{b}.

In arbitrary inner product spaces, \|a\| = (a \cdot a)^{1/2} is the definition of \|a\|, and after one has proven from basic properties of the inner product that |a \cdot b| \leq \|a\|\|b\| one can then define \theta by a \cdot b = \|a\|\|b\| \cos \theta.
 
So you are saying that ||a|| = (a.a)^1/2 is the defination given by the founders and there is no proof for this . But why there is no proof .
 
prashant singh said:
So you are saying that ||a|| = (a.a)^1/2 is the defination given by the founders and there is no proof for this . But why there is no proof .
In general, there is no proof for a definition. That would not be logical.

The definition tells the some meaning of something.
 
  • Like
Likes prashant singh

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
8K
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K